Net capacitance of three identical capacitors in series is 2 micro Farad,What will be their net capacitance if connected in parallel?

Find the ratio of energy stored in the two configurations if they are both connected to the same source?

Let each identical capacitor has capacitance equal to C. When they are connected in series their equivalent capacitance is  Cs=2μf  i.e.1Cs=1C+1C+1C.            or     1Cs=3C       or        C=3 CS       Or      C=3×2μf    or    C=6 μf.Let the source of voltage V is connected to this configuration, therefore the energy stored to this series connection is given by,Us=12Cs V2           Or        Us=12×2×10-6×V2 ------1.Now when these three capacitors are connected in parallel, the net capacitance Cp is given by,Cp=C+C+C.       or     Cp=3C     or   Cp=3×6 μf.           or      Cp=18 μf.Again the energy stored in this configuration,Up=12×Cp×V2             or    Up=12×18×10-6×V2------2.Now comparing the two energies stored given by equations 1 and 2,UsUp=12×2×10-6×V212×18×10-6×V2.            or        UsUp=19.Thus Us : Up =1 : 9.

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Net capacitance of three identical capacitors in series is 2 micro Farad,What will be their net capacitance if connected in parallel?

Find the ratio of energy stored in the two configurations if they are both connecte to the same source?

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 sorry i'm not sure about the energy part, so dont wanna say wrong. hope that helped you. 

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 @darkjuliet : capacitance in series is given by 1/Cs = 1/ C1 + 1/C2 + 1/ C3

since , in this case, capacitors are identical, C1 = C2 = C3. 

Now applying above formula, Cs = C / 3  =====> C = 3 Cs = 3 * 2 uF = 6 uF. 

Capacitance in parallel, Cp = 3 C = 3 * 6 uF = 18 uF. 

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I figured out the energy part. Here it is...

All the three capacitors are connected to the same source and are identical. so C is same for all.

Now formula for energy stored is U = Q2/ 2C

Applying this formula to both Us and Up, and dividing Us / Up,= (Q2/2Cs) / (Q2/2Cp) = Cp / Cs = 18 / 2 = 9 : 1

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